Research

ok thus, $(a-\frac{1}{2})^2\geq 0$ $(b-\frac{1}{2})^2\geq 0$ $\Rightarrow a^2+\frac{1}{4}\geq a$ $b^2+\frac{1}{4}\geq b$ Overall $(a^2+b+\frac{3}{4})(b^2+a+\frac{3}{4})\geq (a+b+\frac{1}{2})(a+b+\frac{1}{2})=(a+b+\frac{1}{2})^2$ $=(a+\frac{1}{4}+b+\frac{1}{4})^2\geq 4(a+\frac{1}{4})(b+\frac{1}{4})=(2a+\frac{1}{2})(2b+\frac{1}{2})$ $\Rightarrow a=b=\frac{1}{2}$ $\rightarrow a+b+2014=\frac{1}{2}+\frac{1}{2}+2014=2015$ That is all